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Find the distance between the points (a) \(\quad(4,-1,2,7)\) and (2,3,1,9) (b) \(\quad(-1,5,-3,2,4)\) and (2,6,2,7,6) (c) \(\quad(5,-2,3,3,1,0)\) and (0,1,5,7,2,1)

Short Answer

Expert verified
Part (a): 5, Part (b): 8, Part (c): 7.48

Step by step solution

01

Understand the Distance Formula in n-Dimensional Space

The distance between two points in n-dimensional space is given by the formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2 + ... + (n_2 - n_1)^2}. \]
02

Calculate the Distance for Part (a)

Identify the coordinates of the two points:\((4, -1, 2, 7)\) and \((2, 3, 1, 9)\).Apply the distance formula:\[ d = \sqrt{(2 - 4)^2 + (3 + 1)^2 + (1 - 2)^2 + (9 - 7)^2} \]\[ d = \sqrt{(-2)^2 + 4^2 + (-1)^2 + 2^2} \]\[ d = \sqrt{4 + 16 + 1 + 4} \]\[ d = \sqrt{25} \]\[ d = 5 \]
03

Calculate the Distance for Part (b)

Identify the coordinates of the two points:\((-1, 5, -3, 2, 4)\) and \((2, 6, 2, 7, 6)\).Apply the distance formula:\[ d = \sqrt{(2 - (-1))^2 + (6 - 5)^2 + (2 - (-3))^2 + (7 - 2)^2 + (6 - 4)^2} \]\[ d = \sqrt{3^2 + 1^2 + 5^2 + 5^2 + 2^2} \]\[ d = \sqrt{9 + 1 + 25 + 25 + 4} \]\[ d = \sqrt{64} \]\[ d = 8 \]
04

Calculate the Distance for Part (c)

Identify the coordinates of the two points:\((5, -2, 3, 3, 1, 0)\) and \((0, 1, 5, 7, 2, 1)\).Apply the distance formula:\[ d = \sqrt{(0 - 5)^2 + (1 + 2)^2 + (5 - 3)^2 + (7 - 3)^2 + (2 - 1)^2 + (1 - 0)^2} \]\[ d = \sqrt{(-5)^2 + 3^2 + 2^2 + 4^2 + 1^2 + 1^2} \]\[ d = \sqrt{25 + 9 + 4 + 16 + 1 + 1} \]\[ d = \sqrt{56} \]\[ d \approx 7.48 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Distance Formula
The distance formula provides a way to calculate the distance between two points. It is a crucial aspect in coordinate geometry.
In two-dimensional space, the distance formula is: \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\).
However, this formula extends to n-dimensional space. To find the distance between two points in n-dimensional space, the formula becomes: \[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + ... + (n_2 - n_1)^2} \].
In this formula:
  • \t- Each pair of coordinates represents a dimension.
  • \t- You subtract the coordinates (like \(x_2 - x_1\)) to find the difference in each dimension.
  • \t- You then square those differences.
  • \t- Sum all squared differences and take the square root of that sum.
This formula is fundamental for calculating distances in any number of dimensions.
n-Dimensional Space
The concept of n-dimensional space extends our three-dimensional understanding to higher dimensions. While our physical world is three-dimensional, mathematically, spaces can have any number of dimensions.
  • Two-Dimensional Space: Defined by coordinates (x, y).
  • Three-Dimensional Space: Defined by coordinates (x, y, z).
  • Four or More Dimensions: Each added coordinate introduces a new dimension. For example, a point in four-dimensional space would have coordinates (x, y, z, w).
In n-dimensional space, points are defined as \(n\)-tuples. This means a point has n coordinates. To visualize higher dimensions, imagine adding new 'directions,' though visualization can be difficult. Real-life applications include data science and machine learning, where datasets can have numerous dimensions.
Coordinate Geometry
Coordinate geometry, or analytic geometry, merges algebra and geometry. It allows geometric problems to be solved using algebraic equations. Points are defined using coordinates, making it easier to understand their geometric relationships.
Key aspects of coordinate geometry include:
  • Coordinates: Each point in space is defined using coordinates (x, y) in two dimensions, (x, y, z) in three dimensions, and so on.
  • Axis and Planes: The coordinate system is typically divided into axes (e.g., x-axis, y-axis) and planes (e.g., xy-plane) in higher dimensions.
  • Distance and Midpoint Formulas: These formulas help calculate distances between points and find midpoints.
Using algebraic methods, we can also derive the equations for geometric shapes like lines, circles, and parabolas.
Euclidean Distance
Euclidean distance is the standard distance measurement in Euclidean space. Named after the Greek mathematician Euclid, it measures the 'straight-line' distance between two points.
The general form of the Euclidean distance between two points \( A = (x_1, y_1, z_1, ...) \) and \( B = (x_2, y_2, z_2, ...) \) is given by: \[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2 + ...} \].
Characteristics of Euclidean distance:
  • It is always positive.
  • It is symmetric, meaning the distance from A to B is the same as from B to A.
  • It satisfies the triangle inequality, implying direct paths are shorter or equal in distance compared to indirect paths.
Euclidean distance is fundamental in various fields like physics, engineering, and computer science. It is used in calculations involving distances on maps, clustering in data analysis, and more.

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Most popular questions from this chapter

Find the eigenvalues and eigenvectors of the following matrices. Do some problems by hand to be sure you understand what the process means. Then check your results by computer. $$\left(\begin{array}{lll} 3 & 2 & 4 \\ 2 & 0 & 2 \\ 4 & 2 & 3 \end{array}\right)$$

Show that if \(D\) is a diagonal matrix, then \(D^{n}\) is the diagonal matrix with elements equal to the \(n^{\text {th }}\) power of the elements of \(\mathrm{D}\). Also show that if \(\mathrm{D}=\mathrm{C}^{-1} \mathrm{MC}\), then \(\mathrm{D}^{n}=\mathrm{C}^{-1} \mathrm{M}^{n} \mathrm{C},\) so \(\mathrm{M}^{n}=\mathrm{CD}^{n} \mathrm{C}^{-1} .\) Hint: For \(n=2,\left(\mathrm{C}^{-1} \mathrm{MC}\right)^{2}=\mathrm{C}^{-1} \mathrm{MCC}^{-1} \mathrm{MC}\) what is \(\mathrm{CC}^{-1} ?\)

Find the symmetric equations and the parametric equations of a line, and/or the equation of the plane satisfying the following given conditions. Plane containing the three points \((0,1,1),(2,1,3),\) and (4,2,1).

Verify that each of the following matrices is Hermitian. Find its eigenvalues and eigenvectors, write a unitary matrix U which diagonalizes \(\mathrm{H}\) by a similarity transformation, and show that \(U^{-1} H U\) is the diagonal matrix of eigenvalues. $$\left(\begin{array}{cc} 2 & i \\ -i & 2 \end{array}\right)$$

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